The smallest interval [a, b] such that int_0^ | vedclass

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(A) Let $I = \int_0^1 \frac{dx}{\sqrt{1 + x^4}}$.For $x \in [0, 1]$,we have $0 \le x^4 \le 1$.Adding $1$ to all sides,we get $1 \le 1 + x^4 \le 2$.Tak

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$[\frac{1}{\sqrt{2}}, 1]$

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(A) Let $I = \int_0^1 \frac{dx}{\sqrt{1 + x^4}}$.For $x \in [0, 1]$,we have $0 \le x^4 \le 1$.Adding $1$ to all sides,we get $1 \le 1 + x^4 \le 2$.Taking the square root,we get $1 \le \sqrt{1 + x^4} \le \sqrt{2}$.Taking the reciprocal,the inequality reverses: $\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + x^4}} \le 1$.Integrating from $0$ to $1$ with respect to $x$,we get $\int_0^1 \frac{1}{\sqrt{2}} dx \le \int_0^1 \frac{dx}{\sqrt{1 + x^4}} \le \int_0^1 1 dx$.This simplifies to $\frac{1}{\sqrt{2}}(1 - 0) \le I \le 1(1 - 0)$,which gives $\frac{1}{\sqrt{2}} \le I \le 1$.Thus,the smallest interval is $[\frac{1}{\sqrt{2}}, 1]$.

The smallest interval $[a, b]$ such that $\int_0^1 \frac{dx}{\sqrt{1 + x^4}} \in [a, b]$ is given by

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